The Poisson
Distribution is based on 1 / e*x times a bunch
of other stuff. The constant e is already not
very intuitive, and 1/e still less so, let
alone 1 / e*x. We met 1 / e
in an earlier problem, Ozzie's
Risk. Here
is another (if slower) way of reaching 1 / e
in a real situation where chance is involved.

Problem

The King's Treasurer, Fernando, each Monday
delivers to Girolamo, the King's Coiner, exactly enough gold to make 100
coins. Girolamo makes the 100 coins, but from one coin he scrapes off
a little gold to provide for his old age. He packs the 100 coins in a
box and sends them each Friday to the King's Inspector, Hernando, who
takes one coin from the box at random and weighs it. If it is light, Girolamo
is executed. If not, nothing happens. Girolamo's appointment as King's
Coiner lasts exactly 100 weeks. What are his chances of surviving his
appointment, to spend his old age savings in peace?

Answer

There are a lot of 100's in this problem,
and numbers that big baffle the intellect. Let's substitute n for 100,
and get a general sense of the problem by asking ourselves what happens
with some smaller, more easily visualized, values of n.

Case 1: n = 1.
Girolamo makes only 1 coin, saves
a little gold from it, and packs it 1 to the box. On inspection, since
it is the only coin in the box, the light coin is chosen and detected,
and Girolamo is executed. His chance of survival is zero.

Case 2: n = 2.
Girolamo makes 2 coins, 1 of which is
light. The chance of that coin being drawn is 1/2, and the chance of it
not being drawn (his chance of surviving
that week) is also 1/2, or 0.5000. His chance of surviving the second week
is exactly the same. To get the chance of his surviving both weeks, we must multiply, whence

(0.5000)
(0.5000) = 0.2500

Case 3: n =
3. Girolamo
makes 3 coins, 1 of which is light. There are thus 2 good coins in the box,
and his chance of not being detected is 2/3 = 0.6667 for each week. Since
there are 3 weeks, his total chance of surviving is

(0.6667)
(0.6667) (0.6667) = 0.2963

General Case.
We can now see the pattern. Girolamo's
chance of surviving for n weeks is a certain fraction times itself n times.
What is that fraction? It is 1 (certainty) minus the chance of getting caught,
and the chance of getting caught is 1/n. To put it in a formula, Girolamo's
chance of surviving is thus

(1 - 1 / n)*n

If we do a few more of these
particular cases, we will see that the chance of survival increases, but
more and more slowly. This is what happens when a number is approaching
some limit. If we list a few more cases, none of which require logarithms
if you manage your 8-place calculator well, we get this picture:

n

p(0 false coins for
n weeks)

1

0.0000

2

0.2500

3

0.2963

4

0.3164

5

0.3277

10

0.3487

20

0.3585

100

0.3660

1000

0.3677

This is about as far as it is convenient
to go with an 8-place calculator and brute mental force. It ought to remind
us of the table at the end of the Ozzie's
Risk problem. That table had already reached (within four decimal
places) the number to which this one is more slowly tending. The number
in question, as can be proved by advanced methods (see Mosteller
42), is 1 / e or approximately 0.3679. This
is Girolamo's Chance. Girolamo's Risk, of not living to
spend his savings, is obviously its complement, or 0.6321.

All things considered, Girolamo would probably
do better to swap jobs with Fernando. Nobody is weighing the gold deliveries,
right?

In general terms, the above series
of values for (1 - 1 / n)*n tends to 1 / e because
(1 - 1 / n)*n itself can be expanded (as powers of 1 / n) to a series whose
terms can, successively, be shown to be equal to

1 / 0! - 1 /
1! + 1 / 2! - 1 / 3! + 1 / 4! - 1 / 5! . . .

which in turn can be shown to
be one of many formulas for 1 / e. (the six terms
given here sum to 0.3667, which is already very close; closer, in fact,
than the sum of 100 terms of the series used above).